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Question Number 141885 by mnjuly1970 last updated on 24/May/21

$$$$$$easyquestion:$$$$if{lim}_{x\to 0}\frac{1-cos(1-cos(1-cos\left(x\right)))}{x^8}=2^a$$$$thena=??$$

Answered by Dwaipayan Shikari last updated on 24/May/21

$$\underset{x\to 0}{\lim }\frac{1-cos(1-cos(1-cos\left(x\right)))}{x^8}=\frac{1-cos(1-cos\frac{x^2}{2})}{x^8}=\frac{1-cos\left(2{sin}^2\frac{x^2}{4}\right)}{x^8}$$$$=\frac{1-cos\left(\frac{x^4}{8}\right)}{x^8}=\frac{2{sin}^2\left(\frac{x^4}{16}\right)}{x^8}=\frac{1}{128}=2^{-7}$$

Commented by mnjuly1970 last updated on 24/May/21

$$thanksalot..$$

Answered by iloveisrael last updated on 24/May/21

$$\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{1-\cos (1-\cos x)}{2}\right)}{x^8}$$$$=\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{2\sin {}^2\left(\frac{1-\cos x}{2}\right)}{2}\right)}{x^8}$$$$=\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{\cancel{2sin}{}^2\left(\frac{\cancel{2sin}{}^2\left(\frac{x}{2}\right)}{\cancel{2}}\right)}{\cancel{2}}\right)}{x^8}$$$$=\underset{x\to 0}{\lim }\frac{2.\sin {}^2\left(\frac{x^4}{2^4}\right)}{x^8}$$$$=\underset{x\to 0}{\lim }\frac{2.\left(\frac{x^8}{2^8}\right)}{x^8}=\frac{1}{2^7}$$

Answered by mathmax by abdo last updated on 24/May/21

$$1-\mathrm{cosx}\sim \frac{{\mathrm{x}}^2}{2}\Rightarrow \cos (1-\mathrm{cosx})\sim \cos \left(\frac{{\mathrm{x}}^2}{2}\right)\sim 1-\frac{{\mathrm{x}}^4}{8}\Rightarrow$$$$1-\cos (1-\mathrm{cosx})\sim \frac{{\mathrm{x}}^4}{8}\Rightarrow \cos (1-\cos (1-\mathrm{cosx}))\sim \cos \left(\frac{{\mathrm{x}}^4}{8}\right)$$$$\sim 1-\frac{{\mathrm{x}}^8}{2.64}\Rightarrow 1-\cos (1-\cos (1-\mathrm{cosx}))\sim \frac{{\mathrm{x}}^8}{2^7}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\frac{1-\cos (1-\cos (1-\mathrm{cosx}))}{{\mathrm{x}}^8}=\frac{1}{128}=2^{-7}\Rightarrow \mathrm{a}=-7$$

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